Desciption:
Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.
For example: Given the below binary tree and sum = 22,
return
代码:
#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#define Elementype int
using namespace std;
int index = 0;
typedef struct TreeNode //树结点
{
Elementype val;
TreeNode *left;
TreeNode *right;
TreeNode(Elementype x) : val(x), left(nullptr), right(nullptr) {}
} *Tree;
//使用先序遍历创建创建二叉树
void MakeBinaryTree(Tree &T, Elementype value[])
{
Elementype c = value[index++];
if (c == '#')
T = nullptr;
else
{
T = new TreeNode(c);
MakeBinaryTree(T->left, value);
MakeBinaryTree(T->right, value);
}
}
//先序遍历
void preOrderTraversal(Tree root)
{
stack<Tree> sk;
Tree p = root;
if (p)
sk.push(p);
while (!sk.empty())
{
p = sk.top();
cout << p->val<<" ";
sk.pop();
if (p->right)
sk.push(p->right);
if (p->left)
sk.push(p->left);
}
}
void pathSum(Tree root, int sum, vector<vector<int>> &result, vector<int> curVec)
{
if (!root)
return;
curVec.push_back(root->val);
if (root->left == nullptr && root->right == nullptr && sum == root->val)
result.push_back(curVec);
pathSum(root->left, sum - root->val, result, curVec);
pathSum(root->right, sum - root->val, result, curVec);
}
vector<vector<int>> pathSum(Tree root, int sum)
{
vector<vector<int>> result; //保存结果
vector<int> curVec; //当前查找的路径
pathSum(root, sum, result, curVec);
return result;
}
int main()
{
Tree root = nullptr; //根结点
Elementype data[] = { 5, 4, 11, 7,'#','#',2,'#','#','#', 8, 13, '#', '#', 4, 5,'#','#', 1,'#','#'};
MakeBinaryTree(root, data);
cout << "先序遍历: ";
preOrderTraversal(root);
cout << endl;
int sum = 22;
vector<vector<int>> path = pathSum(root, sum);
for (auto i : path)
{
cout << "路径有: ";
for (auto j : i)
cout << j << " ";
cout << endl;
}
system("pause");
return 0;
}
测试: